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clear;
clc;
printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 3.1 Page 104 \n') //Example 3.1
// Find Skin Temperature & Aerogel Insulation Thickness
A=1.8; // [m^2] Area for Heat transfer i.e. both surfaces
Ti = 35+273; //[K] - Inside Surface Temperature of Body
Tsurr = 10+273; //[K] - Temperature of surrounding
Tf = 283; //[K] - Temperature of Fluid Flow
e=.95; // Emissivity of Surface
Lst=.003; //[m] - Thickness of Skin
kst=.3; // [W/m.K] Effective Thermal Conductivity of Body
kins = .014; // [W/m.K] Effective Thermal Conductivity of Aerogel Insulation
hr = 5.9; //[W/m^2.k] - Natural Thermal Convectivity from body to air
stfncnstt=5.67*10^(-8); // [W/m^2.K^4] - Stefan Boltzmann Constant
q = 100; //[W] Given Heat rate
//Using Conducion Basic Eq 3.19
Rtot = (Ti-Tsurr)/q;
//Also
//Rtot=Lst/(kst*A) + Lins/(kins*A)+(h*A + hr*A)^-1
//Rtot = 1/A*(Lst/kst + Lins/kins +(1/(h+hr)))
//Thus
//For Air,
h=2; //[W/m^2.k] - Natural Thermal Convectivity from body to air
Lins1 = kins * (A*Rtot - Lst/kst - 1/(h+hr));
//For Water,
h=200; //[W/m^2.k] - Natural Thermal Convectivity from body to air
Lins2 = kins * (A*Rtot - Lst/kst - 1/(h+hr));
Tsa=305; //[K] Body Temperature Assumed
//Temperature of Skin is same in both cases as Heat Rate is same
//q=(kst*A*(Ti-Ts))/Lst
Ts = Ti - q*Lst/(kst*A);
//Also from eqn of effective resistance Rtot F
printf("\n\n (I) In presence of Air, Insulation Thickness = %.1f mm",Lins1*1000)
printf("\n (II) In presence of Water, Insulation Thickness = %.1f mm",Lins2*1000);
printf("\n\n Temperature of Skin = %.2f degC",Ts-273);
//END
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